Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int tu[100 + 5][100 + 5];
int minrow[100 + 5];
int maxrow[100 + 5];
int mincol[100 + 5];
int maxcol[100 + 5];
int tu2[100 + 5][100 + 5];
int main(void) {
  int n, m, i, j, k;
  int flag1;
  int flag2;
  int cnt;
  while (~scanf("%d%d", &n, &m)) {
    queue<pair<int, int> > q;
    memset(tu2, 0, sizeof(tu2));
    flag1 = 0;
    flag2 = 0;
    cnt = 0;
    if (n > m) flag2 = 1;
    for (i = 1; i <= 100 + 5; i++) {
      minrow[i] = mincol[i] = 5000;
      maxrow[i] = maxcol[i] = -5000;
    }
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= m; j++) {
        scanf("%d", &tu[i][j]);
        if (tu[i][j] < minrow[i]) minrow[i] = tu[i][j];
        if (tu[i][j] > maxrow[i]) maxrow[i] = tu[i][j];
        if (tu[i][j] < mincol[j]) mincol[j] = tu[i][j];
        if (tu[i][j] > maxcol[j]) maxcol[j] = tu[i][j];
      }
    }
    if (flag2) {
      for (i = 1; i <= n; i++) tu2[i][1] = mincol[1];
      for (j = 1; j <= mincol[1]; j++) {
        q.push(make_pair(2, 1));
        cnt++;
      }
      for (i = 1; i <= n; i++) {
        int dis;
        if (tu[i][1] > mincol[1]) {
          dis = tu[i][1] - mincol[1];
          for (k = 1; k <= dis; k++) {
            q.push(make_pair(1, i));
            cnt++;
          }
          tu2[i][1] = tu[i][1];
          for (j = 2; j <= m; j++) {
            tu2[i][j] += dis;
            if (tu2[i][j] > tu[i][j]) flag1 = 1;
          }
        }
      }
      int dis;
      for (i = 2; i <= m; i++) {
        if (tu[1][i] < tu2[1][i]) flag1 = 1;
        dis = tu[1][i] - tu2[1][i];
        for (j = 1; j <= dis; j++) {
          q.push(make_pair(2, i));
          cnt++;
        }
        for (k = 1; k <= n; k++) {
          if (tu2[k][i] + dis != tu[k][i]) flag1 = 1;
        }
      }
    } else {
      for (i = 1; i <= m; i++) tu2[1][i] = minrow[1];
      for (j = 1; j <= minrow[1]; j++) {
        q.push(make_pair(1, 1));
        cnt++;
      }
      for (i = 1; i <= m; i++) {
        int dis;
        if (tu[1][i] > minrow[1]) {
          dis = tu[1][i] - minrow[1];
          for (k = 1; k <= dis; k++) {
            q.push(make_pair(2, i));
            cnt++;
          }
          tu2[1][i] = tu[1][i];
          for (j = 2; j <= n; j++) {
            tu2[j][i] += dis;
            if (tu2[j][i] > tu[j][i]) flag1 = 1;
          }
        }
      }
      int dis;
      for (i = 2; i <= n; i++) {
        if (tu[i][1] < tu2[i][1]) flag1 = 1;
        dis = tu[i][1] - tu2[i][1];
        for (j = 1; j <= dis; j++) {
          q.push(make_pair(1, i));
          cnt++;
        }
        for (k = 1; k <= m; k++) {
          if (tu2[i][k] + dis != tu[i][k]) flag1 = 1;
        }
      }
    }
    if (flag1 == 1)
      printf("-1\n");
    else {
      printf("%d\n", cnt);
      while (!q.empty()) {
        if (q.front().first == 1)
          printf("row ");
        else if (q.front().first == 2)
          printf("col ");
        printf("%d\n", q.front().second);
        q.pop();
      }
    }
  }
  return 0;
}